1. UBE529 Resource Allocation

2. 2 Resource Allocation : Schedule The Critical-Section Problem Synchronization Hardware Semaphores Algorithms Distributed Mutual Exclusion Algorithms Central Algorithm Lamport’s Algorithm Ricart-Agrawala Algorithm Suzuki-Kasami Algorithm Raymond’s Algorithm Maekewa’s Algorithm

3. 3 Background Concurrent access to shared data may result in data inconsistency. Maintaining data consistency requires mechanisms to ensure the orderly execution of cooperating processes. Shared-memory solution to bounded-butter problem (Chapter 4) allows at most n – 1 items in buffer at the same time. A solution, where all N buffers are used is not simple. n Suppose that we modify the producer-consumer code by adding a variable counter, initialized to 0 and incremented each time a new item is added to the buffer

4. 4 Bounded-Buffer Shared data #define BUFFER_SIZE 10 typedef struct {... } item; item buffer[BUFFER_SIZE]; int in = 0; int out = 0; int counter = 0;

5. 5 Bounded-Buffer Producer process item nextProduced; while (1) { while (counter == BUFFER_SIZE) ; /* do nothing */ buffer[in] = nextProduced; in = (in + 1) % BUFFER_SIZE; counter++; }

6. 6 Bounded-Buffer Consumer process item nextConsumed; while (1) { while (counter == 0) ; /* do nothing */ nextConsumed = buffer[out]; out = (out + 1) % BUFFER_SIZE; counter--; }

7. 7 Bounded Buffer The statements counter++; counter--; must be performed atomically. Atomic operation means an operation that completes in its entirety without interruption.

8. 8 Bounded Buffer The statement “count++” may be implemented in machine language as: register1 = counter register1 = register1 + 1 counter = register1 The statement “count—” may be implemented as: register2 = counter register2 = register2 – 1 counter = register2

9. 9 Bounded Buffer If both the producer and consumer attempt to update the buffer concurrently, the assembly language statements may get interleaved. Interleaving depends upon how the producer and consumer processes are scheduled. Assume counter is initially 5. One interleaving of statements is: producer: register1 = counter (register1 = 5) producer: register1 = register1 + 1 (register1 = 6) consumer: register2 = counter (register2 = 5) consumer: register2 = register2 – 1 (register2 = 4) producer: counter = register1 (counter = 6) consumer: counter = register2 (counter = 4) The value of count may be either 4 or 6, where the correct result should be 5.

10. 10 Race Condition Race condition: The situation where several processes access – and manipulate shared data concurrently. The final value of the shared data depends upon which process finishes last. To prevent race conditions, concurrent processes must be synchronized.

11. 11 The Critical-Section Problem n processes all competing to use some shared data Each process has a code segment, called critical section, in which the shared data is accessed. Problem – ensure that when one process is executing in its critical section, no other process is allowed to execute in its critical section.

12. 12 Solution to Critical-Section Problem 1.Mutual Exclusion. If process P i is executing in its critical section, then no other processes can be executing in their critical sections. 2.Progress. If no process is executing in its critical section and there exist some processes that wish to enter their critical section, then the selection of the processes that will enter the critical section next cannot be postponed indefinitely. 3.Bounded Waiting. A bound must exist on the number of times that other processes are allowed to enter their critical sections after a process has made a request to enter its critical section and before that request is granted. Assume that each process executes at a nonzero speed No assumption concerning relative speed of the n processes.

13. 13 Software Solutions In addition to mutual exclusion, prevent mutual blocking: 1.Process outside of its CS must not prevent other processes from entering its CS. 2. Process must not be able to repeatedly reenter its CS and starve other processes (fairness) 3. Processes must not block each other forever (deadlock) 4. Processes must not repeatedly yield to each other (“after you”-- “after you” livelock)

14. 14 Shared Memory : Peterson’s Mutual Exclusion Algorithm Use a “ turn ” variable to break a tie: int c1 = 0, c2 = 0, WillWait; cobegin p1: while (1) { c1 = 1; willWait = 1; while (c2 && (WillWait==1)); /*wait*/ CS1; c1 = 0; program1; } //... Guarantees mutual exclusion and no blocking

15. 15 Cooperation Problems with software solutions: n Difficult to program and to verify n Processes loop while waiting (busy-wait) n Applicable to only to critical problem: Competition for a resource Cooperating processes must also synchronize Classic generic scenario: Producer  Buffer  Consumer

16. 16 Bakery Algorithm Before entering its critical section, process receives a number. Holder of the smallest number enters the critical section. If processes P i and P j receive the same number, if i < j, then P i is served first; else P j is served first. The numbering scheme always generates numbers in increasing order of enumeration; i.e., 1,2,3,3,3,3,4,5... Critical section for n processes

17. 17 Bakery Algorithm Notation <  lexicographical order (ticket #, process id #) n (a,b) < c,d) if a < c or if a = c and b < d n max (a 0,…, a n-1 ) is a number, k, such that k  a i for i - 0, …, n – 1 Shared data boolean choosing[n]; int number[n]; Data structures are initialized to false and 0 respectively

18. 18 Bakery Algorithm do { choosing[i] = true; number[i] = max(number[0], number[1], …, number [n – 1])+1; choosing[i] = false; for (j = 0; j < n; j++) { while (choosing[j]) ; while ((number[j] != 0) && (number[j,j] < number[i,i])) ; } critical section number[i] = 0; remainder section } while (1);

19. 19 Synchronization Hardware Test and modify the content of a word atomically boolean TestAndSet(boolean &target) { boolean rv = target; target = true; return rv;} Shared data: boolean lock = false; Process P i do { while (TestAndSet(lock)) ; critical section lock = false; remainder section }

20. 20 Synchronization Hardware Atomically swap two variables. void Swap(boolean &a, boolean &b) { boolean temp = a; a = b; b = temp; } Shared data (initialized to false): boolean lock; boolean waiting[n]; Process P i do { key = true; while (key == true) Swap(lock,key); critical section lock = false; remainder section }

21. 21 Semaphores Synchronization tool that does not require busy waiting. Semaphore S – integer variable can only be accessed via two indivisible (atomic) operations wait (S): while S  0 do no-op; S--; signal (S): S++;

22. 22 Critical Section of n Processes Shared data: semaphore mutex; //initially mutex = 1 Process Pi: do { wait(mutex); critical section signal(mutex); remainder section } while (1);

23. 23 Semaphore Implementation Define a semaphore as a record typedef struct { int value; struct process *L; } semaphore; Assume two simple operations: n block suspends the process that invokes it. n wakeup(P) resumes the execution of a blocked process P.

24. 24 Implementation Semaphore operations now defined as wait(S): S.value--; if (S.value < 0) { add this process to S.L; block; } signal(S): S.value++; if (S.value <= 0) { remove a process P from S.L; wakeup(P); }

25. 25 Semaphore as a General Synchronization Tool Execute B in P j only after A executed in P i Use semaphore flag initialized to 0 Code: P i P j   Await(flag) signal(flag)B

26. 26 Deadlock and Starvation Deadlock – two or more processes are waiting indefinitely for an event that can be caused by only one of the waiting processes. Let S and Q be two semaphores initialized to 1 P 0 P 1 wait(S);wait(Q); wait(Q);wait(S);  signal(S);signal(Q); signal(Q)signal(S); Starvation – indefinite blocking. A process may never be removed from the semaphore queue in which it is suspended.

27. 27 Two Types of Semaphores Counting semaphore – integer value can range over an unrestricted domain. Binary semaphore – integer value can range only between 0 and 1; can be simpler to implement. Can implement a counting semaphore S as a binary semaphore.

28. 28 Classical Problems of Synchronization Bounded-Buffer Problem Readers and Writers Problem Dining-Philosophers Problem

29. 29 Bounded-Buffer Problem Shared data semaphore full, empty, mutex; Initially: full = 0, empty = n, mutex = 1

30. 30 Bounded-Buffer Problem Producer Process do { … produce an item in nextp … wait(empty); wait(mutex); … add nextp to buffer … signal(mutex); signal(full); } while (1);

31. 31 Bounded-Buffer Problem Consumer Process do { wait(full) wait(mutex); … remove an item from buffer to nextc … signal(mutex); signal(empty); … consume the item in nextc … } while (1);

32. 32 Dining-Philosophers Problem Shared data semaphore chopstick[5]; Initially all values are 1

33. 33 Dining-Philosophers Problem Philosopher i: do { wait(chopstick[i]) wait(chopstick[(i+1) % 5]) … eat … signal(chopstick[i]); signal(chopstick[(i+1) % 5]); … think … } while (1);

34. 34 Mutual Exclusion in Distributed Systems Resource sharing Avoiding concurrent update on shared data Controlling the grain of atomicity Medium Access Control in Ethernet Collision avoidance in wireless broadcasts

35. 35 Why mutual exclusion? Some appplications are: Resource sharing Avoiding concurrent update on shared data Controlling the grain of atomicity Medium Access Control in Ethernet Collision avoidance in wireless broadcasts

36. 36 Specifications ME1. At most one process in the CS. (Safety property) ME2. No deadlock. (Safety property) ME3. Every process trying to enter its CS must eventually succeed. This is called progress. (Liveness property) Progress is quantified by the criterion of bounded waiting. It measures fairness by answering the question: Between two consecutive CS trips by one process, how many times other processes can enter the CS? There are many solutions, both on the shared memory model and the message-passing model

37. 37 Centralized Solution clients Client do true  send request; reply received  enter CS; send release; od Server do request received and not busy  send reply; busy:= true request received and busy  enqueue sender release received and queue is empty  busy:= false release received and queue not empty  send reply to the head of the queue od busy: boolean server queue req reply release

38. 38 Comments Centralized solution is simple. Leader maintains a pending queue of events Requests are granted in the order they are received But the server is a single point of failure. This is BAD. Can we do better? Yes!

39. 39 Central MUTEX Algorithm : Client

40. 40 Central Coordinator Algorithm

41. 41 // Centralized mutex algorithm public class CentMutex extends Process implements Lock {... public synchronized void requestCS() { sendMsg(leader, "request"); while (!haveToken) myWait(); } public synchronized void releaseCS() { sendMsg(leader, "release"); haveToken = false; } public synchronized void handleMsg(Msg m, int src, String tag) { if (tag.equals("request")) { if (haveToken){ sendMsg(src, "okay"); haveToken = false; } else pendingQ.add(src); } else if (tag.equals("release")) { if (!pendingQ.isEmpty()) { int pid = pendingQ.removeHead(); sendMsg(pid, "okay"); } else haveToken = true; } else if (tag.equals("okay")) { haveToken = true; notify(); }

42. 42 Decentralized solution 1 : Lamport’s Algorithm 1. Broadcast a timestamped request to all. 2. Request received  enqueue it in local Q. Not in CS  send ack, else postpone sending ack until exit from CS. 3. Enter CS, when (i) You are at the head of your Q (ii) You have received ack from all 4. To exit from the CS, (i) Delete the request from Q, and (ii) Broadcast a timestamped release 5. When a process receives a release message, it removes the sender from its Q.

43. 43 Lamport’s Algorithm program Lamport’s Algorithm define m: msg try, done : boolean Q : queue of msg {Q.i is process i entry} N : integer initially try = false {turns true when a process wants CS} in = false { aprocess enters CS only when in is true} done = false {turns true when a process wants to exit CS} 1. do try -> m := (i, req, t);  j : j !=i send m to j enqueue i in Q try :=false 2.(m.type=request)-> j:=sender enqueue j in Q; send(i,ack, t’) to j

44. 44 Lamport’s Algorithm 3. (m.type = ack)-> N := N + 1 4. (m.type = release)-> j := m.sender; deque Q.j from Q 5. (N = n-1) and (  j != i : Q.j.ts > Q.i.ts) -> in := true { process enters CS} 6. in and done -> in := false; N := 0; dequeue Q.i from Q;  j : j != i send (i, release, t’’) to j; done : = false

45. 45 Lamport’s Algorithm

46. 46 // Lamport’s mutual exclusion algorithm public class LamportMutex extends Process implements Lock { public synchronized void requestCS() { v.tick(); q[myId] = v.getValue(myId); broadcastMsg("request", q[myId]); while (!okayCS()) myWait(); } public synchronized void releaseCS() { q[myId] = Symbols.Infinity; broadcastMsg("release", v.getValue(myId)); } boolean okayCS() { for (int j = 0; j < N; j++){ if(isGreater(q[myId], myId, q[j], j)) return false; if(isGreater(q[myId], myId, v.getValue(j), j))return false; } return true; } public synchronized void handleMsg(Msg m, int src, String tag) { int timeStamp = m.getMessageInt(); v.receiveAction(src, timeStamp); if (tag.equals("request")) { q[src] = timeStamp; sendMsg(src, "ack", v.getValue(myId)); } else if (tag.equals("release")) q[src] = Symbols.Infinity; notify(); // okayCS() may be true now }

47. 47 Analysis of Lamport’s algorithm Can you show that it satisfies all the properties (i.e. ME1, ME2, ME3) of a correct solution? Observation. Any two processes taking a decision must have identical views of their queues. Proof of ME1. At most one process can be in its CS at any time.  j in CS  Qj.ts.j < Qk.ts.k  k in CS  Qk.ts.k < Qj.ts.j Impossible.

48. 48 Analysis of Lamport’s algorithm Proof of FIFO fairness. timestamp of j’s request < timestamp of k’s request implies j enters its CS before k Suppose not. So, k enters its CS before j. So k did not receive j’s request. But k received the ack from j for its own req. This is impossible if the channels are FIFO. Message complexity = 3(N-1) k j Req (20) ack Req (30)

49. 49 Analysis of Lamport’s algorithm Proof of ME2. (No deadlock) The waiting chain is acyclic. i waits for j i is behind j in all queues j does not wait for I Proof of ME3. (progress) New requests join the end of the queues, so new requests do not pass the old ones

50. 50 Analysis of Lamport’s algorithm Proof of FIFO fairness. timestamp (j) < timestamp (k)  j enters its CS before k does so Suppose not. So, k enters its CS before j. So k did not receive j’s request. But k received the ack from j for its own req. This is impossible if the channels are FIFO. (No such guarantee can be given if the channels are not FIFO) Message complexity = 3(N-1) (N-1 requests + N-1 ack + N-1 release) Ensures that processes enter the critical section in the order of timestamps of their requests Requires 3(N-1) messages per invocation of the critical section k j Req (20) ack Req (30)

51. 51 Decentralized algorithm 2 : Ricart-Agrawala’s Algorithm {Ricart & Agrawala’s algorithm} What is new? 1. Broadcast a timestamped request to all. 2. Upon receiving a request, send ack if -You do not want to enter your CS, or -You are trying to enter your CS, but your timestamp is higher than that of the sender. (If you are in CS, then buffer the request) 3. Enter CS, when you receive ack from all. 4. Upon exit from CS, send ack to each pending request before making a new request. (No release message is necessary)

52. 52 Ricart Agrawala Algorithm program Ricart Agrawala Algorithm define m: msg try, want, in : boolean A : array [0..n-1] of boolean t : timestamp N : integer { number of acknowledgements} initially try = false {turns true when a process wants CS} in = false { aprocess enters CS only when in is true} want = false {turns false when a process exits CS} N = 0 A[k] = false { for every k : 0<=k<=n-1} 1. do try -> m := (i, req, t);  j : j !=i send m to j try :=false; want := true

53. 53 RA Algorithm 2a. (m.type=request) AND (~want OR m.ts < t ) -> send(i,ack, t’) to m. Sender 2b. (m.type=request) AND (~want OR m.ts > t ) -> A [sender] := true 3. (m.type = ack)-> N := N + 1 5. (N = n-1) -> in := true { process enters CS} want := false 6. in AND ~want -> in := false; N := 0;  k : send (i, ack, t’) to k;  k : A[k] := false; od

54. 54 Ricart & Agrawala’s algorithm ME1. Prove that at most one process can be in CS. ME3. Prove that FIFO fairness holds even if channels are not FIFO Message complexity = 2(N-1) (N-1 requests + N-1 acks - no release message) k j req ack TS(j) < TS(k)

55. 55 Ricart-Agrawala MUTEX Algorithm

56. 56 public class RAMutex extends Process implements Lock { public synchronized void requestCS() { c.tick(); myts = c.getValue(); broadcastMsg("request", myts); numOkay = 0; while (numOkay < N-1) myWait(); } public synchronized void releaseCS() { myts = Symbols.Infinity; while (!pendingQ.isEmpty()) { int pid = pendingQ.removeHead(); sendMsg(pid, "okay", c.getValue()); } public synchronized void handleMsg(Msg m, int src, String tag) { int timeStamp = m.getMessageInt(); c.receiveAction(src, timeStamp); if (tag.equals("request")) { if ((myts == Symbols.Infinity ) || (timeStamp < myts) ||((timeStamp == myts)&&(src<myId )))//not interested in CS sendMsg(src, "okay", c.getValue()); else pendingQ.add(src); } else if (tag.equals("okay")) { numOkay++; if (numOkay == N - 1) notify(); // okayCS() may be true now } }}

57. 57 Decentralized algorithm 3 : Maekewa’s Algorithm {Maekawa’s algorithm} - First solution with a sublinear O(sqrt N) message complexity. - “Close to” Ricart-Agrawala’s solution, but each process is required to obtain permission from only a subset of peers

58. 58 Maekawa’s algorithm With each process i, associate a subset S i.Divide the set of processes into subsets that satisfy the following two conditions: i  S i  i,j :  i,j  n-1 :: S i  S j ≠  Main idea. Each process i is required to receive permission from S i only. Correctness requires that multiple processes will never receive permission from all members of their respective subsets. 0,1,21,3,5 2,4,5 S0S0 S1S1 S2S2

59. 59 Maekawa’s algorithm Example. Let there be seven processes 0, 1, 2, 3, 4, 5, 6 S 0 ={0, 1, 2} S 1 ={1, 3, 5} S 2 ={2, 4, 5} S 3 ={0, 3, 4} S 4 ={1, 4, 6} S 5 ={0, 5, 6} S 6 ={2, 3, 6}

60. 60 Maekawa’s algorithm Version 1 {Life of process I} 1. Send timestamped request to each process in S i. 2. Request received  send ack to process with the lowest timestamp. Thereafter, "lock" (i.e. commit) yourself to that process, and keep others waiting. 3. Enter CS if you receive ack from each member in S i. 4. To exit CS, send release to every process in S i. 5. Release received  unlock yourself. Then send ack to the next process with the lowest timestamp. S 0 = {0, 1, 2} S 1 = {1, 3, 5} S 2 = {2, 4, 5} S 3 = {0, 3, 4} S 4 = {1, 4, 6} S 5 = {0, 5, 6} S 6 = {2, 3, 6}

61. 61 Maekawa’s algorithm-version 1 ME1. At most one process can enter its critical section at any time. Let i and j attempt to enter their Critical Sections S i   S j ≠   there is a process k  S i   S j Process k will never send ack to both. So it will act as the arbitrator and establishes ME1 S 0 = {0, 1, 2} S 1 = {1, 3, 5} S 2 = {2, 4, 5} S 3 = {0, 3, 4} S 4 = {1, 4, 6} S 5 = {0, 5, 6} S 6 = {2, 3, 6}

62. 62 Maekawa’s algorithm-version 1 ME2. No deadlock Unfortunately deadlock is possible! From S 0 ={0,1,2}, 0,2 send ack to 0, but 1 sends ack to 1; From S 1 ={1,3,5}, 1,3 send ack to 1, but 5 sends ack to 2; Prom S 2 ={2,4,5}, 4,5 send ack to 2, but 2 sends ack to 0; Now, 0 waits for 1, 1 waits for 2, and 2 waits for 0. So deadlock is possible! S 0 = {0, 1, 2} S 1 = {1, 3, 5} S 2 = {2, 4, 5} S 3 = {0, 3, 4} S 4 = {1, 4, 6} S 5 = {0, 5, 6} S 6 = {2, 3, 6}

63. 63 Maekawa’s algorithm-Version 2 Avoiding deadlock If processes receive messages in increasing order of timestamp, then deadlock “could be” avoided. But this is too strong an assumption. Version 2 uses three more messages: - failed - inquire - relinquish S 0 = {0, 1, 2} S 1 = {1, 3, 5} S 2 = {2, 4, 5} S 3 = {0, 3, 4} S 4 = {1, 4, 6} S 5 = {0, 5, 6} S 6 = {2, 3, 6}

64. 64 Maekawa’s algorithm-Version 2 New features in version 2 Send ack and set lock as usual. If lock is set and a request with larger timestamp arrives, send failed (you have no chance). If the incoming request has a lower timestamp, then send inquire (are you in CS?) to the locked process. - Receive inquire and at least one failed message  send relinquish. The recipient resets the lock. S 0 = {0, 1, 2} S 1 = {1, 3, 5} S 2 = {2, 4, 5} S 3 = {0, 3, 4} S 4 = {1, 4, 6} S 5 = {0, 5, 6} S 6 = {2, 3, 6}

65. 65 Maekawa’s algorithm-Version 2

66. 66 Comments Let K = |S i |. Let each process be a member of D subsets. When N = 7, K = D = 3. When K=D, N = K(K-1)+1. So K is of the order √N -The message complexity of Version 1 is 3√N. Maekawa’s analysis of Version 2 reveals a complexity of 7√N Sanders identified a bug in version 2 …

67. 67 Token-passing Algorithms Suzuki-Kasami algorithm Completely connected network of processes There is one token in the network. The owner of the token has the permission to enter CS. Token will move from one process to another based on demand.

68. 68 Suzuki-Kasami Algorithm Process i broadcasts (i, num) Each process maintains -an array req: req[j] denotes the sequence no of the latest request from process j (Some requests will be stale soon) Additionally, the holder of the token maintains -an array last: last[j] denotes the sequence number of the latest visit to CS from for process j. - a queue Q of waiting processes req: array[0..n-1] of integer last: array [0..n-1] of integer Q Sequence number of the request

69. 69 Suzuki-Kasami Algorithm When a process i receives a request (i, num) from process k, it sets req[k] to max(req[k], num) and enqueues the request in its Q When process i sends a token to the head of Q, it sets last[i] := its own num, and passes the array last, as well as the tail of Q, The holder of the token retains process k in its Q only if 1+ last[k] = req[k] This guarantees the freshness of the request Req: array[0..n-1] of integer Last: Array [0..n-1] of integer

70. 70 Suzuki-Kasami’s algorithm {Program of process j} Initially,  i: req[i] = last[i] = 0 * Entry protocol * req[j] := req[j] + 1 Send (j, req[j]) to all Wait until token (Q, last) arrives Critical Section * Exit protocol * last[j] := req[j]  k ≠ j: k  Q  req[k] = last[k] + 1  append k to Q; if Q is not empty  end (tail-of-Q, last) to head-of-Q fi * Upon receiving a request (k, num) * req[k] := max(req[k], num)

71. 71 Suzuki Kasami’s Algorithm

72. 72 public class CircToken extends Process implements Lock { public synchronized void initiate() { if (haveToken) sendToken(); } public synchronized void requestCS() { wantCS = true; while (!haveToken) myWait(); } public synchronized void releaseCS() { wantCS = false; sendToken(); } void sendToken() {... } public synchronized void handleMsg(Msg m, int src, String tag) { if (tag.equals("token")) { haveToken = true; if (wantCS) notify(); else { Util.mySleep(1000); sendToken(); }

73. 73 Example 0 2 1 3 4 req=[1,0,0,0,0] last=[0,0,0,0,0] req=[1,0,0,0,0] initial state

74. 74 Example 0 2 1 3 4 req=[1,1,1,0,0] last=[0,0,0,0,0] req=[0,1,1,0,0] 1 & 2 send requests

75. 75 Example 0 2 1 3 4 req=[1,1,1,0,0] last=[1,0,0,0,0] Q=(1,2) req=[0,1,1,0,0] 0 prepares to exit CS

76. 76 Example 0 2 1 3 4 req=[1,1,1,0,0] req=[0,1,1,0,0] last=[1,0,0,0,0] Q=(2) req=[0,1,1,0,0] 0 passes token (Q and last) to 1

77. 77 Example 0 2 1 3 4 req=[2,1,1,1,0] last=[1,0,0,0,0] Q=(2,0,3) req=[2,1,1,1,0] 0 and 3 send requests

78. 78 Example 0 2 1 3 4 req=[2,1,1,1,0] last=[1,1,0,0,0] Q=(0,3) req=[2,1,1,1,0] 1 sends token to 2

79. 79 Raymond’s tree-based algorithm 1,4 4,7 1 1 4 1,4,7 want to enter their CS

80. 80 Raymond’s Algorithm 1,4 4,7 1 4 2 sends the token to 6

81. 81 Raymond’s Algorithm 4 4,7 4 The message complexity is O(diameter) of the tree. Extensive Empirical measurements show that the average diameter of randomly chosen trees of size n is O(log n). Therefore, the authors claim the average message complexity to be O(log n) 6 forwards the token to 1 4

82. 82 Dining Philosopher Algorithm Eating rule: A process can eat only when it is a source Edge reversal: After eating, reverse orientations of all the outgoing edges

83. 83 Dining Philosphers Algorithm

84. 84 public class DinMutex extends Process implements Lock { public synchronized void requestCS() { myState = hungry; if (haveForks()) myState = eating; else for (int i = 0; i < N; i++) if (request[i] && !fork[i]) { sendMsg(i, "Request"); request[i] = false; } while (myState != eating) myWait(); } public synchronized void releaseCS() { myState = thinking; for (int i = 0; i < N; i++) { dirty[i] = true; if (request[i]) { sendMsg(i, "Fork"); fork[i] = false; } } boolean haveForks() { for (int i = 0; i < N; i++) if (!fork[i]) return false; return true; } public synchronized void handleMsg(Msg m, int src, String tag) { if (tag.equals("Request")) { request[src] = true; if ((myState != eating) && fork[src] && dirty[src]) { sendMsg(src, "Fork"); fork[src] = false; if (myState == hungry){ sendMsg(src, "Request"); request[src] = false; } } else if (tag.equals("Fork")) { fork[src] = true; dirty[src] = false; if (haveForks()) { myState = eating; notify(); } } } }

85. 85 Drinking Philosophers Algorithm